Is the integral in the first display taken with respect to volume measure, or something else? If volume measure, this is trivially false for any manifold $M$ with infinite volume (e.g. $M = \mathbb{R}^n$), since then for any compact $K$, $c(M \setminus K) \ge \operatorname{Vol}(M \setminus K) = \infty$. Or are there other assumptions missing?
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Nate EldredgeJul 13 '13 at 14:35

Also, I cleaned up the formatting in your question. Punctuation marks in sentences such as . , ? ! : ; should always be followed by a space. And when a display equation comes at the end of a sentence, put the period inside the display ($$a+b.$$ not $$a+b$$.), otherwise it appears on the next line and looks bad.
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Nate EldredgeJul 13 '13 at 14:40

I don't understand. $W^{1,2}$ functions are not continuous unless $\dim(M) =1$, so that $c(A)$ does not make sense. Even if you would restrict it to smooth functions, say, then it would be always zero, as you can just take constant functions.
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Matthias LudewigJul 13 '13 at 21:14

@Kofi: note the definition of $c(A)$ only requires $u \ge 1$ almost everywhere on $A$, which is a well-defined statement for $u \in W^{1,2}$. Also, the $W^{1,2}$ norm is usually $\|u\|_{W^{1,2}}^2 = \|u\|_{L^2}^2 + \|\nabla u\|_{L^2}^2$, so that for $u=1$ we get the volume of $M$, not zero. (And if $M$ has infinite volume, constants are not in $W^{1,2}$ at all.)
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Nate EldredgeJul 14 '13 at 3:08

1 Answer
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As discussed in comments, $W^{1,2}$ is interpreted here as $W^{1,2}_0$, the completion of $C^\infty_c(M)$ in the $W^{1,2}$ norm $\|f\|_{W^{1,2}}^2 = \int_M (f^2 + |\nabla f|^2)\,dVol$. Also, the usual definition of capacity involves $\|u\|_{W^{1,2}}^2$ (your version is then the square root of this); I'll use the usual convention, though it doesn't affect the answer.

In general the capacity $c$ need not be tight (my comment was mistaken). Consider the one-dimensional manifold $M = (0,1)$. By Sobolev embedding (which has an elementary proof in this case), each function in $W^{1,2}$ is absolutely continuous and vanishes at the boundary $\{0,1\}$. In particular, if $K$ is compact then there is no $f \in W^{1,2}$ with $f\ge 1$ on $(0,1) \setminus K$, so $c((0,1) \setminus K) = \infty$.

Indeed, $c$ is tight if and only if $1 \in W^{1,2}(M)$.

For the forward direction, recall that since $\varepsilon$ is Dirichlet, if $f \in W^{1,2}$ then $f \wedge 1, f \vee 0 \in W^{1,2}$ as well. If $c$ is tight then in particular there is a compact set $K$ with $c(M \setminus K) < \infty$, i.e. there exists $f \in W^{1,2}(M)$ with $f \ge 1$ a.e. on $M \setminus K$. We can also find a function $g \in W^{1,2}$ with $g \ge 1$ on $K$ (indeed, we could take $g \in C^\infty_c(M)$ by a standard cutoff function construction). Then $1 = ((f \vee 0) + (g \vee 0)) \wedge 1 \in W^{1,2}$.

Conversely, if $1 \in W^{1,2}(M)$ there is a sequence $f_n \in C^\infty_c(M)$ with $f_n \to 1$ in $W^{1,2}$-norm. If $K_n$ is the support of $f_n$ then $g_n := 1 - f_n$ is a $W^{1,2}$ function with $g_n = 1$ on $M \setminus K_n$. So $c(M \setminus K_n) \le \|g\|_{W^{1,2}} = \|1 - f_n\|_{W^{1,2}} \to 0$.